# Adiabatic/Reversible application of external field to Ideal Gas

1. Mar 7, 2009

### a_h

Hello Fellow Physicists!

I'm having trouble with a problem from a graduate course in Statistical Mechanics. It's a two part question; I've got the first part, it's the second part that's giving me trouble.

The problem is about reversible/adiabatic processes, thinking of "reversible" as meaning "constant entropy." In part a (which I got, but it may help to have it here), we need to show that the evolution of a classical ideal gas obeys

$$TV^{\frac{2}{3}}=\mbox{constant.}$$

I did this by using $$S=-\frac{\partial F}{\partial T}$$ with $$F=U-\mu N, U=\frac{3}{2}NT$$. To get $$\frac{\partial \mu}{\partial T}$$, I solved

$$e^{\frac{\mu}{T}}=n\lambda^{3}=\frac{N}{V} \frac{h^{3}}{(2\pi mT)^{3/2}}$$

for mu. ($$\lambda$$ is the thermodynamic wavelength, n is the number density of the gas.)

Now for part b. We have a classical ideal gas in a cylinder. It is in equilibrium at a temperature $$T_{0}$$. Now we slowly (adiabatically and reversibly) apply a uniform external field to the container, in a direction along its axis of symmetry. When we are done, there is a potential in our container given by

$$u(z)=fz$$

where z is the distance along that symmetry axis and f is a constant. The question is, what is the temperature of the gas at the end of the process?

I would think that we should use the same assumptions:

$$S=-\frac{\partial F}{\partial T}=\mbox{constant}$$

and $$F=U-\mu N$$, but I don't know how u(z) gets incorporated into U. Any ideas?

Thanks for all of your time, everyone.